Question

Two point charges are placed over the x-axis as follows: charge q1=+3nC is at x=0.26m, and charge q2=+6nC is at x=-0.45m. Find the magnitude and direction of the total force exerted by these two charges over a negative point charge q3=-8nC that is at x=0m.

Answer 1

Answer:

 F =  5.33 10⁻⁶ N   ,        directed to the left

Explanation:

For this exercise we will use the force addition law where each force is electric given by Coulomb's law

           F = ∑ F

           F = F₁₃ + F₂₃

where the force is given by

           F = k q₁ q₂ / r₁₂²

in our case

           F₁₃ = k q₁ q₃ / r₁₃²

           F₂₃ = k q₂ q₃ / r₂₃²

            F = k q₃ (q₁ / r₁₃² + q₂ / r₂₃²)

as the charges are of the same sign the electric force is repulsive and due to the position of the charges it is directed to the left

let's calculate

           F = 9 10⁹ 8 10⁻⁹ (3 10⁻⁹ / (0.26 - 0) 2 + 6 10⁻⁹ / (0.45-0) 2)

           F = 72 (44,379 + 29,630) 10⁻⁹

           F =   5.33 10⁻⁶ N          

directed to the left